Question

Let ${\alpha }_1,{\beta }_1\left({\alpha }_1<{\beta }_1\right)$ be the roots of the equation $x^2-bx+1=0$ and ${\alpha }_2,{\beta }_2\left({\alpha }_2<{\beta }_2\right)$ be the roots of the equation $a{x}^2+bx+2=0$ such that $\frac{{\alpha }_1+{\alpha }_2}{{\alpha }_1{\alpha }_2}=\frac{{\beta }_1+{\beta }_2}{{\beta }_1{\beta }_2}=1$.

List I		List II
P	The value of a equals	1	-4
Q	The value of $b$ equals	2	1
R	The value of $\left({\beta }_1+2{\alpha }_2\right)$ equals	3	3
S	The value of $\left({\alpha }_1-2{\alpha }_2\right)$ equals	4	4

• A. P-1, Q-4, R-3, S-2
• B. P-2, Q-3, R-1, S-4
• C. P-1, Q-1, R-3, S-4
• D. P-1, Q-3, R-2, S-1

Answer 1

Answer: (A)

Given, $\frac{1}{{\alpha }_1}+\frac{1}{{\alpha }_2}=1$ and $\frac{1}{{\beta }_1}+\frac{1}{{\beta }_2}=1$
$\text{ Add }\Rightarrow \left(\frac{1}{{\alpha }_1}+\frac{1}{{\alpha }_2}\right)+\left(\frac{1}{{\beta }_1}+\frac{1}{{\beta }_2}\right)=2\Rightarrow \left(\frac{1}{{\alpha }_1}+\frac{1}{{\beta }_1}\right)+\left(\frac{1}{{\alpha }_2}+\frac{1}{{\beta }_2}\right)=2\Rightarrow \frac{{\alpha }_1+{\beta }_1}{{\alpha }_1{\beta }_1}+\frac{{\alpha }_2+{\beta }_2}{{\alpha }_2{\beta }_2}=2$
$\Rightarrow \frac{b}{1}+\frac{-b/a}{2/a}=2\Rightarrow b-\frac{b}{2}=2\Rightarrow b=4$
Hence, roots of the first equation are $2+\sqrt{3},2-\sqrt{3}$
$\therefore {\alpha }_1=2-\sqrt{3}\text{ and }{\beta }_1=2+\sqrt{3}$
Also second equation is $a{x}^2+4x+2=0$.
Now, $\frac{1}{{\alpha }_1}+\frac{1}{{\alpha }_2}=1$
$\therefore \frac{1}{{\alpha }_2}=1-\frac{1}{{\alpha }_1}\Rightarrow \frac{1}{{\alpha }_2}=1-(2+\sqrt{3})=-1-\sqrt{3}\Rightarrow {\alpha }_2=\frac{-1}{1+\sqrt{3}}$
Put ${\alpha }_2$ in second equation we get $a=-4.$